2018-G12-Math-E

version: 1.1CHAPTER1 FUNCTIONS AND LIMITS Animation 1.1: Function Machine Source and credit: eLearn.Punjab

11.. FQuunacdtiroantsicaEnqduLaimtioitnss eLearn.Punjab eLearn.Punjab1.1 INTRODUCTION Functions are important tools by which we describe the real world in mathematicalterms. They are used to explain the relationship between variable quantities and hence playa central role in the study of calculus.1.1.1 Concept of Function The term function was recognized by a German Mathematician Leibniz (1646 - 1716)to describe the dependence of one quantity on another. The following examples illustrateshow this term is used:(i) The area “A” of a square depends on one of its sides “x” by the formula A = x2, sowe say that A is a function of x.(ii) The volume “ V ” of a sphere depends on its radius “r” by the formula V = 4 pr3, so 3we say that V is a function of r. A function is a rule or correspondence, relating two sets in such a way that eachelement in the irst set corresponds to one and only one element in the second set. Thus in, (i) above, a square of a given side has only one area. And in, (ii) above, a sphere of a given radius has only one volume.Now we have a formal deinition:1.1.2 Deinition (Function - Domain - Range) A Function f from a set X to a set Y is a rule or a correspondence that assigns to eachelement x in X a unique element y in Y. The set X is called the domain of f.The set of corresponding elements y in Y is called the range of f. Unless stated to the contrary, we shall assume hereafter that the set X and Y consist ofreal numbers. version: 1.1 2

11.. FQuunacdtiroantsicaEndquLaimtioitnss eLearn.Punjab eLearn.Punjab1.1.3 Notation and Value of a Function If a variable y depends on a variable x in such a way that each value of x determinesexactly one value of y, then we say that “y is a function of x”. Swiss mathematician Euler (1707-1783) invented a symbolic way to write the statement“y is a function of x” as y = f(x) , which is read as “y is equal to f of x”.Note: Functions are often denoted by the letters such as f, g, h , F, G, H and so on. A function can be thought as acomputing machine f that takes an input x,operates on it in some way, and producesexactly one output f(x). This output f(x) iscalled the value of f at x or image of x underf. The output f(x) is denoted by a singleletter, say y, and we write y = f(x). The variable x is called the independent variable of f, and the variable y is calledthe dependent variable of f. For now onward we shall only consider the function inwhich the variables are real numbers and we say that f is a real valued function of realnumbers.Example 1: Given f(x) = x3 - 2x2 + 4x - 1, ind(i) f(0) (ii) f(1) (iii) f(-2) (iv) f(1 + x) (v) f(1/x), x ≠ 0Solution: f(x) = x3 - 2x2 + 4x - 1 (i) f(0) = 0 - 0 + 0 - 1 = - 1 (i) f(1) = (1)3 - 2(1)2 + 4(1) - 1 = 1 - 2 + 4 -1 = 2 (ii) f(-2) = (- 2)3 - 2 (-2 )2 + 4 (-2) - 1 = - 8 - 8 - 8 - 1 = -2 5 (iii) f(1 + x) = (1 + x)3 - 2(1 + x)2 + 4(1 + x) - 1 = 1 + 3x + 3x2 + x3 - 2 - 4x - 2x2 + 4 + 4x - 1 = x3 + x2 + 3x + 2 version: 1.1 3

11.. FQuunacdtiroantsicaEnqduLaimtioitnss eLearn.Punjab eLearn.Punjab(iv) f(1/x) = (1/x)3 - 2(1/x)2 + 4 (1/x) - 1 = 1 - 2 + 4 - 1, x ≠ 0 x3 x2 xExample 2: Let f(x) = x2. Find the domain and range of f.Solution: f(x) is deined for every real number x. Further for every real number x, f(x) = x2 is a non-negative real number. So Domain f = Set of all real numbers. Range f = Set of all non-negative real numbers.Example 3: Let f(x) = x . Find the domain and range of f. x2 - 4Solution: At x = 2 and x = -2, f(x) = x 4 is not deined. So Domain x2 - except -2 and 2 . f = Set of all real numbers Range f = Set of all real numbers.Example 4: Let f(x) = x2 - 9 . Find the domain and range of f.Solution: We see that if x is in the interval -3 < x < 3, a square root of a negative number isobtained. Hence no real number y = x2 - 9 exists. So Domain f = { x d R : |x| 8 3 } = (-T, -3] j [3, + T) Range f = set of all positive real numbers = (0, + T)1.1.4 Graphs of Algebraic functions If f is a real-valued function of real numbers, then the graph of f in the xy-plane isdeined to be the graph of the equation y = f(x). The graph of a function f is the set of points {(x, y)| y = f(x)} , x is in the domain of f in theCartesian plane for which (x, y) is an ordered pair of f. The graph provides a visual techniquefor determining whether the set of points represents a function or not. If a vertical lineintersects a graph in more than one point, it is not the graph of a function. version: 1.1 4

11.. FQuunacdtiroantsicaEndquLaimtioitnss eLearn.Punjab Explanation is given in the igure. eLearn.PunjabMethod to draw the graph:To draw the graph of y = f(x), we give arbitrary values of our choice to x and ind thecorresponding values of y. In this way we get ordered pairs (x1 , y1) , (x2 , y2), (x3 , y3) etc. Theseordered pairs represent points of the graph in the Cartesian plane. We add these points andjoin them together to get the graph of the function.Example 5: Find the domain and range of the function f(x) = x2 + 1 and draw its graph.Solution: Here y = f(x) = x2 + 1 We see that f(x) = x2 +1 is deined for every real number. Further, for every real numberx, y = f(x) = x2 + 1 is a non-negative real number. Hence Domain f = set of all real numbers and Range f = set of all non-negative real numbers except the points 0 7 y < 1. For graph of f(x) = x2 +1, we assign some values to x from its domain and ind thecorresponding values in the range f as shown in the table: x -3 -2 -1 0 1 2 3 y = f(x) 10 5 2 1 2 5 10 Plotting the points (x, y) and joining them with a smooth curve,we get the graph of the function f(x) = x2 + 1, which is shown in theigure. version: 1.1 5

11.. FQuunacdtiroantsicaEnqduLaimtioitnss eLearn.Punjab eLearn.Punjab1.1.5 Graph of Functions Deined Piece-wise. When the function f is deined by two rules, we draw the graphs of two functions asexplained in the following example:Example 7: Find the domain and range of the function deined by:[f(x) =x when 0 7x 71 also draw its graph. x - 1 when 1 < x 72Solution: Here domain f = [0, 1] j [1, 2] = [0, 2]. This function is composed of the followingtwo functions: (ii) f(x) = x - 1 , when 1 < x 7 2(i) f(x) = x when 0 7 x 7 1To ind th table of values of x and y = f(x) in each case, we take suitable values to x in thedomain f. Thus Table of y = f(x) = x - 1: Table for y = f(x) = xx 0 0.5 0.8 1 x 1.1 1.5 1.8 2y = f(x) 0 0.5 0.8 1 y = f(x) 0.1 0.5 0.8 1Plotting the points (x, y) and joining them we gettwo straight lines as shown in the igure. This isthe graph of the given function.1.2 TYPES OF FUNCTIONS Some important types of functions are given below:1.2.1 Algebraic Functions Algebraic functions are those functions which are deined by algebraic expressions.We classify algebraic functions as follows: version: 1.1 6

11.. FQuunacdtiroantsicaEndquLaimtioitnss eLearn.Punjab eLearn.Punjab(i) Polynomial FunctionA function P of the form P(x) = an xn + an-1 xn-1 + an-2 xn-2 + .... + a2 x2 + a1 x + a0for all x, where the coeicient an, an-1, an-2, .... , a2, a1, a0 are real numbers and the exponentsare non-negative integers, is called a polynomial function.The domain and range of P(x) are, in general, subsets of real numbers. If an ≠ 0 , then P(x) is called a polynomial function of degree n and an is the leadingcoeicient of P(x) . For example, P(x) = 2x4 - 3x3 + 2x - 1 is a polynomial function of degree 4 with leadingcoeicient 2.(ii) Linear Function If the degree of a polynomial function is 1, then it is called a linear function. A linearfunction is of the form: f(x) = ax + b (a ≠ 0), a, b are real numbers. For example f(x) = 3x + 4 or y = 3x + 4 is a linear function. Its domain and range are theset of real numbers.(iii) Identity Function For any set X, a function I : X \" X of the form I(x) = x \" x d X , is called an identityfunction. Its domain and range is the set X itself. In particular, if X = R , then I(x) = x , for all xd R , is the identity function.(iv) Constant function Let X and Y be sets of real numbers. A function C : X \" Y deined by C(x) = a , \" x d X , ad Y and ixed, is called a constant function. For example, C : R \" R deined by C(x) = 2, \" x d R is a constant function.(v) Rational Function P(x)Q(x) ≠A0f,uisnccatilolendRa(xr)aotfitohneaflofrumncQt(ixo)n,.where both P(x) and Q(x) are polynomial functions andThe domain of a rational function R(x) is the set of all real numbers x for which Q(x) ≠ 0.1.2.2 Trigonometric FunctionsWe denote and deine trigonometric functions as follows: version: 1.1 7

11.. FQuunacdtiroantsicaEnqduLaimtioitnss eLearn.Punjab eLearn.Punjab(i) y = sin x, Domain = R, Range -1 7 y 7 1.(ii) y = cos x , Domain = R, Range -1 7 y 7 1.(iii) y = tan x, Domain = {x : xdR and x = (2n + 1) 2p, n an integer}, Range = R(iv) y = cot x, Domain = {x : xdR and x ≠ np, n an integer}, Range= R p,(v) y = sec x, Domain = {x : xdR and x ≠ (2n + 1) 2 n an integer}, Range= R(vi) y = csc x, Domain = {x : xdR and x ≠ np, n an integer}, Range = y 8 1, y 7 -11.2.3 Inverse Trigonometric FunctionsWe denote and deine inverse trigonometric functions as follows:(i) y = sin-1 x ⇔ x = sin y, where - p ≤ y ≤ p , -1 ≤ x ≤ 1 22(ii) y = cos-1 x ⇔ x = cos y, where 0 ≤ y ≤ p, -1 ≤ x ≤ 1(iii) y = tan-1 x ⇔ x = tan y, where - p < y < p , - ∞ < x < ∞ 221.2.4 Exponential Function A function, in which the variable appears as exponent (power), iscalled an exponential function. The functions, y = eax, y = ex, y = 2x =ex ln 2, etc are exponential functions of x.1.2.5 Logarithmic Function If x = ay , then y = loga x , where a > 0, a ≠ 1 is called Logarithmic Function of x.(i) If a = 10, then we have log10 x (written as lg x) which is known as the common logarithm of x.(ii) If a = e, then we have loge x (written as In x) which is known as the natural logarithm of x. version: 1.1 8

11.. FQuunacdtiroantsicaEndquLaimtioitnss eLearn.Punjab eLearn.Punjab1.2.6 Hyperbolic Functions(i) sinh x = 1 (ex - e-x) is called hyperbolic sine function. Its domain and range are 2 the set of all real numbers.(ii) cosh x = 1 (ex + e-x) is called hyperbolic cosine function. Its domain is the set of 2 all real numbers and the range is the set of all numbers in the interval [1, +T)(iii) The remaining four hyperbolic functions are deined in terms of the hyperbolic sine and the hyperbolic cosine function as follows:==tanh x = csoinshh xx ex - e-x sech x = 1 2 ex + e-x cosh x ex + e-x==coth x = csoinshh xx ex + e-x csch x = 1 x ex - e-x sinh 2 ex - e-xThe hyperbolic functions have same properties that resemble to those oftrigonometric functions.1.2.7 Inverse Hyperbolic Functions The inverse hyperbolic functions are expressed in terms of natural logarithms and weshall study them in higher classes.(i) sinh-1 x = ln(x + x2 + 1 ), for all x (iv) coth -1 x= 1 ln  x +1  , x <1 2 x -1(ii) cosh-1 x = ln(x + x2 - 1 ) x ≥ 1 (v) sech -1 x = ln  1 + 1 - x2  , 0 < x ≤1 x x(iii) tanh-1 x= 1 ln  1+ x  , x <1 (vi) csch -1 x = ln  1 + 1 + x2  , x≠0 2 1- x x x1.2.8 Explicit Function If y is easily expressed in terms of the independent variable x, then y is called an explicitfunction of x. For example (i) y = x2 + 2x - 1 (ii) =y x -1 are explicit functions of x. version: 1.1 9

11.. FQuunacdtiroantsicaEnqduLaimtioitnss eLearn.Punjab eLearn.PunjabSymbolically it can be written as y = f(x).1.2.9 Implicit Function If x and y are so mixed up and y cannot be expressed in terms of the independentvariable x, then y is called an implicit function of x. For example,(i) x2 + xy + y2 = 2 (ii) xy 2 -y+9 =1 are implicit functions of x and y. xySymbolically it is written as f(x, y) = 0.(ix) Parametric Functions Some times a curve is described by expressing both x and y as function of a thirdvariable “t” or “q” which is called a parameter. The equations of the type x = f(t) and y = g(t)are called the parametric equations of the curve . The functions of the form: x = at2 x = a cos t (iii) x = a cos q (iv) x = a sec q (i) (ii) y = a sin t y = b sin q y = a tan q y = at are called parametric functions. Here the variable t or q is called parameter.1.2.10 Even Function A function f is said to be even if f(-x) = f(x) , for every number x in the domain of f.For example: f(x) = x2 and f(x) = cos x are even functions of x.Here f(-x) = (-x)2 = x2 = f(x) and f(-x) = cos (-x) = cos x = f(x)1.2.11 Odd Function A function f is said to be odd if f(-x) = -f(x) , for every number x in the domain of f.For example, f(x) = x3 and f(x) = sin x are odd functions of x. Here f(-x) = (-x)3 = -x3 = -f(x) and f(-x) = sin(-x) = -sin x = -f(x) version: 1.1 10

11.. FQuunacdtiroantsicaEndquLaimtioitnss eLearn.Punjab eLearn.PunjabNote : In both the cases, for each x in the domain of f, -x must also be in the domain of f.Example 1: Show that the parametric equations x = a cos t and y = a sin t representthe equation of the circle x2 + y2 = a2Solution: The parametric equations are x = a cos t (i) y = a sin t (ii)We eliminate the parameter “t” from equations (i) and (ii).By squaring we get, x2 = a2 cos2 t y2 = a2 sin2 tBy adding we get, x2 + y2 = a2 cos2 t + a2 sin2 t = a2 (cos2 t + sin2 t) ∴ x2 + y2 = a2, which is equation of the circle.Example 2: Prove the identities(i) cosh2 x - sinh2 x = 1 (ii) cosh2 x + sinh2 x = cosh 2xSolution: We know that sinh x = ex - e-x (1) 2 and cosh x = ex + e-x (2) 2Squaring (1) and (2) we haveNow (i) sinh2 x = e2x + e-2x - 2 and cosh2 x = e2x + e-2x + 2 ∴ 44 cosh2 x - sinh2 x = e2x + e-2x + 2 - e2x + e-2x - 2 44 = e2x + e-2x + 2 - e2x - e-2x + 2 = 4 44 cosh2 x - sinh2 x = 1 version: 1.1 11

11.. FQuunacdtiroantsicaEnqduLaimtioitnss eLearn.Punjaband (ii) cosh2 x + sinh2 x = e2 x + e-2x + 2 + e2 x + e-2x - 2 eLearn.Punjab 4 4 version: 1.1 = e2 x + e-2x + 2 + e2x + e-2x - 2 4 2e2x + 2e-2x e2x + e-2x == 42 ∴ cosh2 x + sinh2 x = cosh 2xExample 3: Determine whether the following functions are even or odd.(a) f(x) = 3x4 - 2x2 + 7 (b) f (x) = 3x (c) f(x) = sin x + cos x x2 + 1Solution: f(-x) = 3(-x)4 - 2(-x)2 + 7 = 3x4 - 2x2 + 7 = f(x) (a) f(x) = 3x4 - 2x2 + 7 is even. Thus f ( - x) = 3( - x) 1 - 3x 1 = - f (x) (b) ( - x)2 + x2 + Thus f (x) = 3x is odd (c) x2 + 1 f(-x) = sin(-x) + cos(-x) = -sin x + cos x ≠ ± f(x) Thus f(x) = sin x + cos x is neither even nor odd EXERCISE 1.11. Given that: (a) f(x) = x2 - x (b) f (x) = x + 4 Find (i) f(-2) (ii) f(0) (iii) f(x - 1) (iv) f(x2 + 4)2. Find f(a + h) - f(a) and simplify where, h (i) f(x) = 6x - 9 (ii) f(x) = sin x (iii) f(x) = x3 + 2x2 - 1 (iv) f(x) = cos x 12

11.. FQuunacdtiroantsicaEndquLaimtioitnss eLearn.Punjab eLearn.Punjab3. Express the following: (a) The perimeter P of square as a function of its area A. (b) The area A of a circle as a function of its circumference C. (c) The volume V of a cube as a function of the area A of its base.4. Find the domain and the range of the function g deined below, and(i) g(x) = 2x - 5 (ii) g(x) = x2 - 4(iii) g(x) = x + 1 (iv) g(x) = x - 3(v) g(x) = 64x-+37x , x ≤ -2 (vi) g(x) = 2xx-+11 , x<3 , -2 < x , 3≤x(vii) g(x) = x2 - 3x + 2 , x ≠ -1 (viii) g(x) = x2 - 16 , x ≠ 4 x+1 x -45. Given f(x) = x3 - ax2 + bx + 1 If f(2) = -3 and f(-1) = 0 . Find the values of a and b.6. A stone falls from a height of 60m on the ground, the height h afterx seconds is approximately given by h(x) = 40 - 10x2(i) What is the height of the stone when:. (a) x = 1 sec ? (b) x = 1.5 sec ? (c) x = 1.7 sec ?(ii) When does the stone strike the ground?7. Show that the parametric equations:(i) x = at2 , y = 2at represent the equation of parabola y2 = 4ax(ii) x = acosq , y = bsinq represent the equation of ellipse x2 + y2 =1 a2 b2(iii) x = asecq , y = btanq represent the equation of hyperbola x2 y2 (ii) sech2 x = 1 - tanh2 x a2 - b2 = 18. Prove the identities: (i) sinh 2x = 2sinh x cosh x (iii) csch2 x = coth2 x - 1 version: 1.1 13

11.. FQuunacdtiroantsicaEnqduLaimtioitnss eLearn.Punjab eLearn.Punjab9. Determine whether the given function f is even or odd.(i) f(x) = x3 + x (ii) f(x) = (x + 2)2(iii) f (x) = x x2 + 5 (iv) f (x) = x- 1, x ≠ -1 x+ 1(v) f (x) = x2 3 + 6 (vi) f (x) = x3 - x x2 + 11.3 COMPOSITION OF FUNCTIONS AND INVERSE OF AFUNCTION Let f be a function from set X to set Y and g be a function from set Y to set Z. Thecomposition of f and g is a function, denoted by gof, from X to Z and is deined by (gof)(x) = g(f(x)) = gf(x) , for all xdX.1.3.1 Composition of Functions Remember That: Briely we write gof as gf.ExplanationConsider two real valued functions f and g deined byf(x) = 2x + 3 and g(x) = x2then gof(x) = g(f (x) ) = g(2x + 3) = (2x + 3)2The arrow diagram of two consecutive mappings, ffollowed by g, denoted by gf is shown in the igure.Thus a single composite function gf(x) is equivalentto two successive functions f followed by g.Example 1: Let the real valued functions f and g be deined by f(x) = 2x + 1 and g(x) = x2 - 1Obtain the expressions for (i) fg (x) (ii) gf (x) (iii) f2 (x) (iv) g2 (x) version: 1.1 14

11.. FQuunacdtiroantsicaEndquLaimtioitnss eLearn.Punjab eLearn.PunjabSolution:(i) fg (x) = f (g (x)) = f ( x2 - 1) = 2 (x2 - 1) +1 = 2x2 - 1(ii) gf (x) = g (f(x)) = g (2x + 1) = (2x + 1)2 - 1 = 4x2 + 4x(iii) f2(x) = f (f(x)) = f (2x + 1) = 2(2x +1) + 1 = 4x + 3(iv) g2(x) = g(gx) = g (x2 - 1) = (x2 - 1)2 - 1 = x4 - 2x2We observe from (i) and (ii) that fg (x) ≠ gf(x)Note:1. It is important to note that, in general, gf (x) ≠ fg (x) , because gf (x)means that f is applied irst then followed by g, whereas fg (x) means that g is applied irst then followed by f.2. We usually write f as f 2 and ff as f 3 and so on.1.3.2 Inverse of a Function Let f be a one-one function from X onto Y. The inverse function of f denoted by f -1, isa function from Y onto X and is deined by:x = f -1(y), [ y d Y if and only if y = f(x) , [ x d X.Illustration by arrow diagramThe inverse function reverses the correspondenceof the original function, so that f -1(y) = x, when f(x) = yand f(x) = y , when f -1(y) = xWe can ind the composition of the functions f andf -1 as follows: (f -1 of)(x) = f -1(f (x)) = f -1(y) = x and (fof -1)(y) = f (f -1(y)) = f(x) = y We note that f -1 of and fof -1 are identity mappings on the domain and range of f andf -1 respectively.1.3.3 Algebraic Method to ind the Inverse Function The inverse function can be found by using the algebraic method as explained in thefollowing example: version: 1.1 15

11.. FQuunacdtiroantsicaEnqduLaimtioitnss eLearn.Punjab eLearn.PunjabExample 2: Let f : R \" R be the function deined by f(x) = 2x + 1. Find f -1(x)Remember that:The change of name of variable in the deinition of function does not change that functionwhere the domain and range coincide.Solution: We ind the inverse of f as follows: Write f(x) = 2x + 1 = y So that y is the image of x under f. Now solve this equation for x as follows: y = 2x +1 ⇒ 2x = y - 1⇒ x= y-1 2∴ f -1 (y) = 1 ( y - 1) ∴ x = f -1(y) 2To ind f -1(x), replace y by x.∴ f -1(x) = 1 (x - 1) 2Veriication: ( )f  1 - 1)   1 - 1) f -1(x) = f 2 (x = 2 2 (x + 1 = x and f -1 ( f (x)) = f -1 (2x + 1) = 1 (2x + 1 - 1) = x 2Example 3: Without inding the inverse, state the domain and range of f -1, where f (x) = 2 + x -1Solution: We see that f is not deined when x < 1. x -1 varies over the interval [0, +T). ∴ Domain f = [1, +T) As a varies over the interval [1, +T), the value of version: 1.1 16

11.. FQuunacdtiroantsicaEndquLaimtioitnss eLearn.Punjab eLearn.PunjabSo the value of f(x) = 2 + x -1 varies over the interval [2, +T). Therefore range f = [2, +T) By deinition of inverse function f -1, we have domain f -1 = range f = [2, +T) and range f -1 = domain f = [1, +T) EXERCISE 1.21. The real valued functions f and g are deined below. Find(a) fog (x) (b) gof (x) (c) fof (x) (d) gog (x)(i) f(x) = 2x + 1 ; g(x) = 3 , x ≠ 1 x -1(ii) f (x) = x+1 ; g(x) = 1 , x≠ 0(iii) f (x) = 1 , x ≠ 1 ; x2 ; x -1 g(x) = (x2 + 1)2(iv) f(x) = 3x4 - 2x2 g (x) = 2 , x ≠ 0 x2. For the real valued function, f deined below, ind(a) f -1(x) (b) f -1(-1) and verify f (f -1 (x)) = f -1 f(x)) = x(i) f(x) = -2x + 8 (ii) f(x) = 3x3 + 7(iii) f(x) = (-x + 9)3 (iv) f (x) = 2x + 1 , x > 1 x -13. Without inding the inverse, state the domain and range of f -1.(i) f (x) = x + 2 (iii) f (x) = 1 , x ≠ -3 x+3(ii) f (x) = x - 1 , x ≠ 4 (iv) f(x) = (x - 5)2 , x 8 5 x -4 version: 1.1 17

11.. FQuunacdtiroantsicaEnqduLaimtioitnss eLearn.Punjab eLearn.Punjab1.4 LIMIT OF A FUNCTION AND THEOREMS ON LIMITS The concept of limit of a function is the basis on which the structure of calculus rests.Before the deinition of the limit of a function, it is essential to have a clear understanding ofthe meaning of the following phrases:1.4.1 Meaning of the Phrase “x approaches zero”Suppose a variable x assumes in succession a series of values as1, 1 , 1 , 1 , 1 ,... i.e., 1, 1 , 1 , 1 , 1 , ... , 1 ,... 2 4 8 16 2 22 23 24 2nWe notice that x is becoming smaller and smaller as n increases and can be made as smallas we please by taking n suiciently large. This unending decrease of x is symbolically writtenas x \" 0 and is read as “x approaches zero” or “x tends to zero”.Note: The symbol x \" 0 is quite diferent from x = 0 (i) x \" 0 means that x is very close to zero but not actually zero. (ii) x = 0 means that x is actually zero.1.4.2 Meaning of the Phrase “x approaches ininity” Suppose a variable x assumes in succession a series of values as 1 ,10 ,100 ,1000 ,10000 .... i.e., 1,10,102,103.......,10n,... It is clear that x is becoming larger and larger as n increases and can be made as largeas we please by taking n suiciently large. This unending increase of x is symbolically writtenas “x \"T” and is read as “x approaches ininity” or “x tends to ininity”.1.4.3 Meaning of the Phrase “x approaches a”Symbolically it is written as “x \" a” which means that x is suiciently close to but diferentfrom the number a, from both the left and right sides of a i.e; x - a becomes smaller andsmaller as we please but x - a ≠ 0. version: 1.1 18

11.. FQuunacdtiroantsicaEndquLaimtioitnss eLearn.Punjab eLearn.Punjab1.4.4 Concept of Limit of a Function (i) By inding the area of circumscribing regular polygon Consider a circle of unit radius which circumscribes a square (4-sided regular polygon)as shown in igure (1). The side of square is 2 and its area is 2 square unit. It is clear that the area of inscribed4-sided polygon is less than the area of the circum-circle. Bisecting the arcs between the vertices of the square, we get an inscribed 8-sidedpolygon as shown in igure 2. Its area is 2 2 square unit which is closer to the area ofcircum-circle. A further similar bisection of the arcs gives an inscribed 16-sided polygon asshown in igure (3) with area 3.061 square unit which is more closer to the area of circum-circle. It follows that as ‘n’ , the number of sides of the inscribed polygonincreases, the area of polygon increases and becoming nearer to3.142 which is the area of circle of unit radius i.e., pr2 = p(1)2= p c 3.1 42. We express this situation by saying that the limiting value of the area o f the inscribedpolygon is the area of the circle as n approaches ininity, i.e., Area of inscribed polygon \" Area of circle as n \"T Thus area of circle of unit radius = p = 3.142 (approx.)(ii) Numerical Approach Consider the function f(x) = x3 The domain of f(x) is the set of all real numbers. Let us ind the limit of f(x) = x3 as x approaches 2. version: 1.1 19

11.. FQuunacdtiroantsicaEnqduLaimtioitnss eLearn.Punjab eLearn.Punjab The table of values of f(x) for diferent values of x as x approaches 2 from left andright is as follows: from left of 2 2 from right of 2x 1 1.5 1.8 1.9 1.99 1.999 1.9999 2.0001 2.001 2.01 2.1 2.2 2.5 3f(x)=x3 1 3.375 5.832 6.859 7.8806 7.988 7.9988 8.0012 8.012 8.1206 9.261 10.648 15.625 27 The table shows that, as x gets closer and closer to 2 (suiciently close to 2), fromboth sides, f(x) gets closer and closer to 8. We say that 8 is the limit of f(x) when x approaches 2 and is written as: f ( x) → 8 as x → 2 or lim ( x3 ) = 8 x→21.4.5 Limit of a Function Let a function f(x) be deined in an open interval near the number “a” (need not beat a). If, as x approaches “a” from both left and right side of “a”, f(x) approaches a speciicnumber “L” then “L”, is called the limit of f(x) as x approaches a.Symbolically it is written as: Lim f ( x) = L read as “limit of f(x), as x \" a , is L”. x→a It is neither desirable nor practicable to ind the limit of a function by numericalapproach. We must be able to evaluate a limit in some mechanical way. The theorems onlimits will serve this purpose. Their proofs will be discussed in higher classes.1.4.6 Theorems on Limits of Functions Let f and g be two functions, for which Lim f (x) = L and Lim g(x) = M , then x→a x→aTheorem 1: The limit of the sum of two functions is equal to the sum of their limits. Lim  f ( x) + g( x) = Lim f (x) + Lim g(x) =L+M x→a x→a x→aFor example, (Lim x + 5) = Lim x+ Lim 5 =1+5=6 x→1 x→1 x→1 version: 1.1 20

11.. FQuunacdtiroantsicaEndquLaimtioitnss eLearn.Punjab eLearn.PunjabTheorem 2: The limit of the diference of two functions is equal to the diference of their limits.Lim  f ( x) - g( x) = Lim f (x) - Lim g(x) =L-M x→a x→a (x - 5) x→aFor example, Lim = Lim x - Lim 5=3-5=-2 x→3 x→3 x→3Theorem 3: If k is any real number, thenLim kf ( x) =k Lim f (x) = kL x→a x→a (3x)For example: Lim = 3Lim (x ) = 3 (2 ) = 6 x→2 x→2Theorem 4: The limit of the product of the functions is equal to the product of their limits.Lim  f ( x) g( x) =  Lim f ( x)  Lim g( x) = LM x→a x→a x→aFor example: Lim (2x)(x + 4) =  Lim ( 2 x )  Lim ( x + 4) =(2)(5)=10 x→1 x→1 x→1Theorem 5: The limit of the quotient of the functions is equal to the quotient of their limits provided the limit of the denominator is non-zero.Lim  f (x)  = Lim f (x) = L, g( x) ≠ 0, M ≠ 0 g (x) g(x) M x→a x→a Lim x→aFor example: Lim  3x +4  = Lim (3x + 4) = 6+4 = 10 =2 x +3 2+3 5 x→2 x→2 Lim (x + 3) x→2Theorem 6: Limit of  f ( x)n , where n is an integer( )Lim n x→a = Ln  f ( x)n = Lim f (x) x→a ( )Lim 3 x→4 = (5)3For example: (2x - 3)3 = Lim (2x - 3) = 125 x→4 We conclude from the theorems on limits that limits are evaluated by merelysubstituting the number that x approaches into the function. version: 1.1 21

11.. FQuunacdtiroantsicaEnqduLaimtioitnss eLearn.Punjab eLearn.PunjabExample : If P(x) = anxn + an-1 x n-1 + .... + a1 x + a0 is a polynomial function of degree n,then show that Lim P(x) = P(c) x→cSolution: Using the theorems on limits, we have =Lim P=(x) Lim+ (an xn an-1 x+n-1 + .... a1x + a0 x→c x→c = an Lim xn + an-1 Lim xn-1 + .... + a1 Lim x + Lim a0 x→c x→c x→c x→c = an cn + an-1 cn-1 + .... + a1c + a0∴ Lim P(x) = P(c) x→c1.5 LIMITS OF IMPORTANT FUNCTIONS If, by substituting the number that x approaches into the function, we get  0  , then weevaluate the limit as follows: 0 We simplify the given function by using algebraic technique of making factors if possibleand cancel the common factors. The method is explained in the following important limits.1.5.1 Lim xn - an = nan-1 where n is an integer and a > 0 x - a x→aCase 1: Suppose n is a positive integer. By substituting x = a , we get  0  form. So we make factors as follows: 0 xn - an = (x - a) (xn-1 + axn-2 + a2 xn-2 + .... + an-1)( )∴ Lim xn - an = Lim (x - a ) axn-1 + axn-2 a2 xn-3 + . . . . + an-1 x-a x→a x→a x-a = Lim (xn-1 + axn-2 + a2 xn-3 + .... + an-1) (polynomial function) x→a = an-1 + a.an-2 + a2.an-3 + .... + an-1 = an-1 + an-1 + an-1 + .... + an-1 (n terms) = nan-1 version: 1.1 22

11.. FQuunacdtiroantsicaEndquLaimtioitnss eLearn.Punjab eLearn.PunjabCase II: Suppose n is a negative integer (say n = -m) , where m is a positive integer. Now xn - an = x-m - a-m x-a x-a = -1  xm - am  (a ≠ 0) xm am x - a ∴ Lim xn - an = Lim  -1   xm - am  x -a xm am x- a x→a x→a= -1 .( mam-1 ), (By case 1) am am = -ma-m-1 ∴ Lim xn - an = nan-1 (n = - m) x→a x - a1.5.2 L→im + a a = By substituting x = 0, we have  0  form, so rationalizing the numerator. 0 ∴ Lim x+a - a = Lim  x+a - a  x+a + a  x→0 x x x+a + a x→0 = Lim x + a - a x→0 x( x + a + a ) = Lim x x→0 x( x + a + a ) = Lim 1 x→0 x + a + a = 1 =1 a+ a 2 a version: 1.1 23

11.. FQuunacdtiroantsicaEnqduLaimtioitnss eLearn.Punjab eLearn.PunjabExample 1: Evaluate (i) Lim x2 - 1 (ii) Lim x - 3 x2 - x x→3 x - 3 x→1Solution: (i) Lim x2 - 1  0  form (By making factors) x2 - x 0 x→1∴ Lim x2 -1 = Lim ( x - 1)( x + 1) = Lim x+1 = 1+1 =2 x2 -x x( x - 1) x→1 x1 x→1 x→1(ii) Lim x -3  0  form (By making factors of x - 3) x- 3 0 x→3∴ Lim x - 3 = Lim ( x + 3 )( x - 3) x→3 x - 3 x→3 x- 3 = Lim ( x + 3) x→3 = ( 3 + 3) =2 31.5.3 Limit at IninityWe have studied the limits of the functions f(x), f(x) g(x) and f(x) when x \" c (a number) g(x),Let us see what happens to the limit of the function f(x) if c is +T or -T (limits at ininity)i.e. when x \" +T and x \" -T.(a) Limit as x \" +T Let f (x) = 1 , when x ≠ 0 x This function has the property that the value of f(x) can be made as close as we pleaseto zero when the number x is suiciently large.We express this phenomenon by writing Lim 1 = 0 x→∞ x version: 1.1 24

11.. FQuunacdtiroantsicaEndquLaimtioitnss eLearn.Punjab eLearn.Punjab(b) Limit as x \" -T. This type of limits are handled in the same way as limits as x \" +T.i.e.=Lim 1 0, where x ≠ 0 xx→-∞ The following theorem is useful for evaluating limit at ininity.Theorem: Let p be a positive rational number. If xp is deined, then Lim a = 0 and Lim a = 0 ,where a is any real number. xp x→-∞ xp x→+∞For example, L im 6 =0, L im -5 = Lim -5 =0 x3 x x→-∞ x1/ 2 x→±∞ x→-∞ and L im 1 = Lim 1 =0 5x x→+∞ x→+∞ 1 x51.5.4 Method for Evaluating the Limits at Ininity In this case we irst divide each term of both the numerator and the denominator bythe highest power of x that appears in the denominator and then use the above theorem.Example 2: Evaluate Lim 5x4 - 10x2 + 1Solution: -3x3 + 10x2 + 50 x→+∞ Dividing up and down by x3 , we getL im 5x4 - 10x2 +1 = Li 5x - 10/x + 1/x3 = ∞-0+0 = ∞ -3 3 + 10x2 + 50 -3 + 10/x + 50/ 3 -3 + 0 + 0x→+∞ x→+∞Example 3: Evaluate Lim 4x4 - 5x3Solution: 3x5 + 2x2 + 1 x→-∞ Since x < 0, so dividing up and down by (-x)5 = -x5, we getL im 4x4 - 5x3 = Lim -4 / x + 5/x2 = 0+0 =0 3x5 + 2x2 + 1 -3 - 2/x3 - 1/x5 -3 - 0 - 0x→-∞ x→-∞ version: 1.1 25

11.. FQuunacdtiroantsicaEnqduLaimtioitnss eLearn.PunjabExample 4: Evaluate eLearn.PunjabSolution: version: 1.1 (i) Lim 2 - 3x (ii) Lim 2 - 3x x→-∞ 3 + 4x2 x→+∞ 3 + 4x2 (i) Here x2 = x = -x as x < 0 ∴ Dividing up and down by -x, we get Lim 2 - 3x = Lim -2/x + 3 = 0 + 3 = 3 x→-∞ 3 + 4x2 x→-∞ 3 / x2 + 4 0+4 2 (ii) Here x == x ==--xx aas xx > 0 ∴ Dividing up and down by x, we get Lim 2 - 3x = Lim 2/x + 3 = 0 - 3 = -3 x→+∞ 3 + 4x2 x→+∞ 3 / x2 + 4 0+4 21.5.5 xL→im+∞ 1 + 1 n = e. n By the Binomial theorem, we have 1 + 1 n = 1 + n  1  + n( n - 1 )  1 2 + n( n - 1 )( n - 2 )  1 3 +... n n 2 ! n 3! n = 1 +1 + 1 1 - 1  + 1 1 - 1 1 - 2  +... 2! n 3! n n when n → ∞, 1 , 2 , 3 , . . . all tend to zero. nnn∴ L im 1 + 1 n = 1 + 1 + 1 + 1 + 1 + 1 +... n 2! 3! 4! 5! x→∞ = 1 + 1 + 0.5 + 0.166667 + 0.0416667 + ... = 2.718281 ...As approximate value of e is = 2.718281.∴ Lim 1 + 1 n = e . x→+∞ n 26

11.. FQuunacdtiroantsicaEndquLaimtioitnss eLearn.PunjabDeduction Lx→im0 (1 + x)1/ x = e eLearn.Punjab version: 1.1We know that Lim 1 + 1 n = e (i) n in (i) x→∞ put n = 1 , then 1 = x xn When x → 0, n → ∞ As Lim 1 + 1 n = e n x→∞ ∴ Lim (1 + )x 1/ x = e x→01.5.6 Lx→im0 a x- 1 = loge a x Put ax - 1 = y (i) then ax = 1 + y So x = loga (1 + y) From (i) when x \" 0, y \" 0∴ Lim ax - 1 y = Lim 1 x→0 = Lim loga (1 + y) y→0 1 loga (1 + y) x y→0 y = Lim 11 = log ( ) Lim(1 + y)1/ y = e = e a y→0 y→0 loga (1 + y)1/ y loga eDeduction xL→im0  e x- 1  = loge e = 1. xWe know that Lim ax - 1 = loge a (1) x x→0 27

11.. FQuunacdtiroantsicaEnqduLaimtioitnss eLearn.PunjabPut a = e in (1), we have eLearn.Punjab version: 1.1Lim ex - 1 = loge e = 1. x x→0 Important Results to Remember (i) Lim (ex ) = ∞ (ii) Lim (ex ) = Lim  1  = 0, x→∞ e-x x→-∞ x→-∞ (iii) Lim  a  = 0 , where a is any real number. x x→±∞Example 5: Express each limit in terms of the number ‘e’ (a) Lim 1 + 3 2n 1 n n→+∞ (b) Lim (1+2h)h h→0Solution: (a) Observe the resemblance of the limit withLim 1 + 1 n = e nn→∞ 1 + n  6 = 1 + n  6 3 3 1 + 3 2n = 3  1  n n n / 3 1 + 2 n 1 m  6  put m = n/3 , when n ∴ Lim 3 = Lim + 1 = e6 m →∞ n m →∞ n→+∞ m→+∞ 1(b) Observe the resemblance of the limit with Lim (1 + x) x = e , x→0∴ Lim 1 = Lim (1 + 1 2 (put m = 2h, when h → 0, m → 0 h→0  (1 + 2h)h h→0 2h) 2 h (1 1  2 = Lim + m) m = e2 m→0 28

11.. FQuunacdtiroantsicaEndquLaimtioitnss eLearn.Punjab eLearn.Punjab1.5.7 The Sandwitch Theorem Let f, g and h be functions such that f(x) 7 g(x) 7 h(x) for all numbers x in some openinterval containing “c”, except possibly at c itself. If Lim f (x) = L and Lim h(x) = L, then Lim g(x) = L x→c x→c x→cMany limit problems arise that cannot be directly evaluated by algebraic techniques. Theyrequire geometric arguments, so we evaluate an important theorem.1.5.8 If q is measured in radian, then Lim sinq =1 q q →0Proof: To evaluate this limit, we apply a new technique. Take q a positive acute central angleof a circle with radius r = 1. As shown in the igure, OAB represents a sector of the circle. Given OA = OB = 1 (radii of unit circle)∴ In rt ∆OCB, sinq = BC = BC ( OB = 1) OBIn rt ∆OAD, tanq = AD = AD ( OA = 1) OA In terms of q, the areas are expressed as:Produce OB to D so that AD ⊥ OA. Draw BC ⊥ OA. Join AB (i) Area of ∆OAB = 1 OA BC = 1 (1)(sinq ) = 1 sinq (ii) 22 2and (iii) Area of sector OAB = 1 r2q = 1 (1)(q ) = 1q ( r = 1) 22 2 Area of ∆OAD = 1 OA AD = 1 (1)(tanq ) = 1 tanq 22 2 From the igure we see that Area of ∆OAB < Area of sector OAB < Area of ∆OAD ⇒ 1 sinq < q < 1 tanq 2 2 2 As sin q is positive, so on division by 1 sin q, we get 2 version: 1.1 29

11.. FQuunacdtiroantsicaEnqduLaimtioitnss eLearn.Punjab eLearn.Punjab 1< q < 1  0 < q < p  sinq cosq 2 i.e., 1> sinq > cosq or cosq < sinq <1 q q when q \" 0, cos q \" 1Since Sin q is sandwitched between 1 and a quantity approaching 1 itself. qSo, by the sandwitch theorem, it must also approach 1. i.e., lim sinq =1 q q →0Note: The same result holds for -p/2 < q < qExample 6: Evaluate: lim sin 7q q →0 qSolution: Observe the resemblance of the limit with lim sinq =1 q →0 q Let x = 7q so that q = x/7 when q \" 0 , we have x \" 0 ∴ Lim sin 7q = Lim sin x = 7 Lim sin x = (7)(1) = 7 q x→0 x/7 x→0 x q →0Example 7: Evaluate: L im 1- cosq q q →0Solution: 1- cosq = 1- cosq .1 + cosq q q 1 + cosq = 1 - cos2 q = sin2 q = sinq  sinq   1 + 1  q cosq q (1+ cosq ) q (1+ cosq ) 1 - cos q sinq 1 ∴ lim q = lim sinq lim q lim 1+ cosq q →0 q →0 q →0 q →0 = (0)(1)( 1 1 ) 1 + = (0) version: 1.1 30

11.. FQuunacdtiroantsicaEndquLaimtioitnss eLearn.Punjab eLearn.Punjab EXERCISE 1.31. Evaluate each limit by using theorems of limits:(i) Lim (2x + 4) (ii) Lim (3x2 - 2x + 4) (iii) Lim x2 + x + 4 x→3 x→1 x→3(iv) Lim x2 - 4 (v) Lim ( x3 + 1 - x2 + 5 ) (vi) Lim 2x3 + 5x x→2 x→2 x→-2 3x - 22. Evaluate each limit by using algebraic techniques.(i) Lim x3 - x (ii) Lim  3x3 + 4x  (iii) Lim x3 - 8 x→-1 x + 1 x2 + x x→2 x2 + x - 6 x→0(iv) Lim x3 - 3x2 + 3x -1 (v) Lim  x3 + x2  (vi) Lim 2x2 - 32 x3 - x x2 -1 x→4 x3 - 4x2 x→1 x→-1(vii) Lim x- 2 (viii) Lim x + h - x (ix) Lim xn - an x→2 x -2 h→0 h x→a xm - am3. Evaluate the following limits(i) Lim sin7x (ii) Lim sin x0 (iii) Lim 1 - cosq x→0 x x→0 x sinq q →0(iv) Lim sin x (v) Lim sina x (vi) Lim x p -x x→0 sinbx x→0 tan x x→p(vii) Lim 1- cos 2x (viii) Lim 1- cos x (ix) Lim sin2 q x2 sin2 x q x→0 x→0 q →0(x) Lim sec x - cos x (xi) Lim 1 - cos pq (xii) Lim tanq - sinq x→0 x 1 - cos qq sin3 q q →0 q →04. Express each limit in terms of e: 1 2n n 1 n 1  2(i) L im + 1 (ii) L im + 1 (iii) L im - 1 n n n n→+∞ n→+∞ n→+∞(iv) L im 1 + 1 n (v) L im 1 + 4 n (vi) L im (1 + 3x ) 2 3n n x n→+∞ n→+∞ x→0 version: 1.1 31

11.. FQuunacdtiroantsicaEnqduLaimtioitnss eLearn.Punjab eLearn.Punjab ( )1 (1 ) 1  x  x h +(vii) Lim 1 + 2x2 x2 (viii) L im - 2h (ix) L im x→0 1 x h→0 x→∞(x) L im e1/ x -1 , x < 0 (xi) L im e1/ x - 1 , x > 0 e1/ x +1 e1/ x + 1 x→0 x→01.6 Continuous and Discontinuous Functions1.6.1 One-Sided Limits In deining Lim f (x) , we restricted x to an open interval containing c i.e., we studied x→cthe behavior of f on both sides of c. However, in some cases it is necessary to investigateone-sided limits i.e., the left hand limit and the right hand limit.(i) The Left Hand Limit Lim f (x) = L is read as the limit of f(x) is equal to L as x approaches c from the left i.e., x→cfor all x suiciently close to c, but less than c, the value of f(x) can be made as close as weplease to L.(ii) The Right Hand Limit Lim f (x) = M is read as the limit of f(x) is equal to M as x approaches c from the right x→ci.e., for all x suiciently close to c, but greater than c, the value of f(x) can be made as close aswe please to M.Note: The rules for calculating the left-hand and the right-hand limits are the same aswe studied to calculate limits in the preceding section.1.6.2 Criterion for Existence of Limit of a FunctionL=im f (x) L if and o=nly if Lx→i=mc- f (x) L im f (x) L x→c x→c+ version: 1.1 32

11.. FQuunacdtiroantsicaEndquLaimtioitnss eLearn.Punjab eLearn.PunjabExample 1: Determine whether Lim f (x) and Lim f (x) exist, when x→2 x→4 f ( x) = 72x-+ 1 if 0≤ x ≤2 x if 2≤ x ≤4 x if 4≤ x ≤6Solution:(i) L im f (x) = L im (2x +1) =4 +1 = 5 x→2- x→2- Lim f (x) =Lim(7 - x) =7 - 2 = 5 x→2+ x→2+ Since Lx→=im2- f (x) Lx=→im2+ f (x) 5 ⇒ Lim f (x) exists and is equal to 5. x→2(ii) L im f (x) =Lx→im4- (7 - x) =7 - 4 = 3 x→4- Lim=f (x) L=im(x) 4 x→4+ x→4+Since L im f (x) ≠ L im f (x) x→4- x→4+Therefore Lim f (x) does not exist. x→4We have seen that sometimes Lim f (x) = f (c) and sometimes it does not and also sometimes x→cf (c) is not even deined whereas Lim f (x) exists. x→c1.6.3 Continuity of a function at a number (a) Continuous Function A function f is said to be continuous at a number “c” if and only if the following threeconditions are satisied: (i) f (c) is deined. (ii) Lim f (x) exists, (iii) Lim f (x) = f (c) x→c x→c (b) Discontinuous Function If one or more of these three conditions fail to hold at “c”, then the function f is said tobe discontinuous at “c”. version: 1.1 33

11.. FQuunacdtiroantsicaEnqduLaimtioitnss eLearn.Punjab eLearn.PunjabExample 2: Consider the function f(x)= x2 -1 x -1Solution: Here f (1) is not deined ⇒ f (x) is discontinuous at 1. Further Lim f ( x ) = lim x2 - 1 = lim (x + 1) = 2 (finite) x→1 x→1 x - 1 x→1 Therefore f (x) is continuous at any other number x ≠ 1Example 3: For f (x) = 3x2 - 5x + 4, discuss continuity of f at x = 1Solution: Lim f (x) =Lim (3x2 - 5x + 4-) =3 =5 + 4 2. x→1 x→1 ⇒ ∴ and f(1) = 3 - 5 + 4 = 2Example 4: Lim f (x) = f (1) x→1 f (x) is continuous at x = 1 Discuss the continuity of the function f (x) and g (x) at x = 3.(a) f(x)=  x2 -9 if x ≠ 3 (b) g( x) = x2 - 9 if x ≠ 3 x -3 x -3  6 if x = 3Solution: (a) Given f (3) = 6∴ the function f is deined at x = 3.Now Lim f( x ) = Lim x2 - 9 x→3 x→3 x - 3 = Lim ( x + 3)( x - 3) x→3 x - 3 (x + 3) = Lim =6 x→3 As Lim f( x ) = 6 = f( 3 ) x→3∴ f (x) is continuous at x = 3 version: 1.1 34

11.. FQuunacdtiroantsicaEndquLaimtioitnss eLearn.Punjab It is noted that there is no break in the graph. (See igure (i)) eLearn.Punjab (b) g( x ) = x2 - 9 if x ≠ 3 x -3 As g (x) is not deined at x = 3 ⇒ g (x) is discontinuous at x = 3 (See igure (ii)). It is noted that there is a break in the graph at x = 3Example 5: Discuss continuity of f at 3, when f (x) = 2x -1 , if x<3 x +1 , if 3≤xSolution: A sketch of the graph of f is shown in the igure (iii). We see that there is a break in the graph at the point when x = 3 Now f (3) = 2(3) + 1 = 7 ⇒ Condition (i) is satisied. Lim f (x) = Lim f (x - 1) = 3 - 1 = 2 x→3- x→3- L im f (x) = L im f (2x + 1) =6 +1=7 x→3+ x→3+ L im f (x)≠≠ L im f (x) →x →- 3- →x →+ 3+ i.e. condition (ii) is not satisied ∴ Lim f (x) does not exist x→3 Hence f(x) is not continuous at x = 3 EXERCISE 1.41. Determine the left hand limit and the right hand limit and then, ind the limit of the following functions when x \" c (i) f(x) = 2x2 + x - 5, c = 1 (ii) f( x ) = x2 - 9 , c= -3 (iii) f( x ) = x - 5 , c = 5 x - 3 version: 1.1 35

11.. FQuunacdtiroantsicaEnqduLaimtioitnss eLearn.Punjab2. Discuss the continuity of f(x) at x = c: eLearn.Punjab version: 1.1 (i) f (x) = 24 x + 5 if x≤2 x + 1 if ,c =2 x2 (ii) f (x) = 3x -1 if x<1 4 if x = 1, c = 1 if x >1 2x3. If f (x) =  3x 1 if x ≤ -2 x2 - if -2<x <2  3 Discuss continuity if x ≥2 x = -2 at x =2 and4. If f (x) = cx +2 , x ≤ -1 find \"c\" so that Lim f (x) exists. +2 , , x→-1 x > -15. Find the values m and n, so that given function f is continuous at x = 3. (i) f (x) = -m2nxx + 9 if x<3 (ii) f (x) = mxx2 if x<3 if x=3 if x≥3 if x>36. If f ( x ) =  2x + 5 - x + 7 , x≠ 2 x -2 x=2 k, Find value of k so that f is continuous at x = 2. 36

11.. FQuunacdtiroantsicaEndquLaimtioitnss eLearn.Punjab eLearn.Punjab1.7 Graphs We now learn the method to draw the graphs of the Explicit Functions like y = f(x) ,where f(x) = ax, ex, loga x , and loge x.1.7.1 Graph of the Exponential Function f(x) = ax Let us draw the graph of y = 2x, here a = 2. We prepare the following table for diferent values of x and f(x) near the origin: x -4 -3 -2 -1 0 1 2 3 4 y = f(x) = 2x 0.0625 0.125 0.25 0.5 1 2 4 8 16 Plotting the points (x, y) and joining them with smoothcurve as shown in the igure, we get the graph of y = 2x. From the graph of 2x the characteristics of the graphof y = ax are observed as follows:If a > 1, (i) ax is always +ve for all real values of x. (ii) ax increases as x increases. (iii) ax = 1 when x = 0 (iv) ax \" 0 as x \"-T1.7.2 Graph of the Exponential Function f(x) = ex As the approximate value of ‘e’ is 2.718 The graph of ex has the samecharacteristics and properties as that of ax whena > 1 (discussed above). We prepare the table of some values of x and f(x)near the origin as follows: version: 1.1 37

11.. FQuunacdtiroantsicaEnqduLaimtioitnss eLearn.Punjab eLearn.Punjab x -3 -2 -1 01 2 3 y = f(x) = ex 0.05 0.135 0.36 1 2.718 7.38 20.07 Plotting the points (x, y) and joining them with smooth curve as shown, we get thegraph of y = ex.1.7.3 Graph of Common Logarithmic Function f(x) = lg x. If x = 10y, then y = lg x Now for all real values of y, 10y > 0 ⇒ x > 0 This means lg x exists only when x > 0 ⇒ Domain of the lg x is +ve real numbers. Note: lg x is undeined at x = 0. For graph of f(x) = lg x, we ind the values of lg x fromthe common logarithmic table for various values of x > 0. Table of some of the corresponding values of x and f(x) is as under: x \"0 0.1 1 2 4 6 8 10 \"+T y = f(x) = lg x \"-T -1 0 0.30 0.60 0.77 0.90 1 \"+T Plotting the points (x, y) and joining them with a smooth curve we get the graph asshown in the igure.1.7.4 Graphs of Natural Logarithmic Function f(x) = In x: The graph of f(x) = In x has similar properties as thatof the graph of f(x) = lg x. By using the table of natural logarithm for various valuesof x, we get the graph of y = In x as shown in the igure. version: 1.1 38

11.. FQuunacdtiroantsicaEndquLaimtioitnss eLearn.Punjab eLearn.Punjab1.7.5 Graphs of Implicit Functions(a) Graph of the circle of the form x2 + y2 = a2Example 1: Graph the circle x2 + y2 = 4 (1)Solution: The graph of the equation x2 + y2 = 4 is a circle of radius 2, centered at theorigin and hence there are vertical lines that cut the graph more than once. This can also beseen algebraically by solving (1) for y in terms of x. y = ± 4 - x2 The equation does not deine y as a function of x.For example, if x = 1, then y = ± 3 .Hence ( (1, 3) ) and ( (1, - 3) ) are two points on the circle and vertical line passes throughthese two points.We can regard the circle as the union of two semi-circles. y = 4 - x2 and y = - 4 - x2Each of which deines y as a function of x.We observe that if we replace (x, y) in turn by (-x, y), (x, -y) and (-x, -y), there is nochange in the given equation. Hence the graph is symmetric with respect to the y-axis, x-axisand the origin. y2 = 4 ⇒ y = ±2 x = 0 impliesx = 1 implies y2 = 3 ⇒ y = ± 3x = 2 implies y2 = 0 ⇒ y = 0By assigning values of x, we ind the values of y. So we prepare a table for some valuesof x and y satisfying equation (1). version: 1.1 39

11.. FQuunacdtiroantsicaEnqduLaimtioitnss eLearn.Punjab eLearn.Punjab x01 3 2 -1 - 3 -2 y ±2 ± 3 ±1 0 ± 3 ±1 0 Plotting the points (x , y) and connecting them with a smooth curve as shown in theigure, we get the graph of a circle.(b) The graph of ellipse of the form x2 + y2 =1 a2 b2Example 2: Graph x2 + y2 = 1 i.e., 9x2 + 4y2 = 36 22 32Solution: We observe that if we replace (x, y) in turn by (-x, y),(x,- y) and (-x, -y) , there is no change in the given equation. Hence thegraph is symmetric with respect to the y-axis, x-axis and the origin. y = 0 implies x2 = 4 ⇒ x = ±2 x = 0 implies y2 = 9 ⇒ y = ±3Therefore x-intercepts are 2 and -2 and y-intercepts are 3 and -3 By assigning values of x, we ind the values of y. So we preparea table for some values of x and y satisfying equation (1). x 0 1 2 -1 -2 y ±3 ± 27 0 ± 27 0 44Ploting the points (x, y), connecting these points with a smooth curve as shown in theigure, we get the graph of an ellipse.1.7.5 Graph of parametric Equations(a) Graph the curve that has the parametric equationsx = t2 , y = t -2 7 t 7 2 (3) version: 1.1 40

11.. FQuunacdtiroantsicaEndquLaimtioitnss eLearn.Punjab eLearn.PunjabSolution: For the choice of t in [-2, 2], we prepare a table forsome values of x and y satisfying the given equation. t -2 -1 0 1 2 x41014 y -2 -1 0 1 2 We plot the points (x, y) , connecting thesepoints with a smooth curve shown in igure, weobtain the graph of a parabola with equationy2 = x.1.7.6 Graphs of Discontinuous FunctionsExample 1: Graph the function deined by y =  xx - 1 when 0 ≤ x ≤1 when 1 < x ≤2Solution: The domain of the function is 0 7 x 7 2For 0 7 x 7 1, the graph of the function is that of y = xand for 1 < x 7 2 , the graph of the function is that of y = x - 1We prepare the table for some values of x and y in 0 7 x 7 2 satisfying the equations y= x and y = x - 1x 0 0.5 0.8 1 1.5 1.8 2y 0 0.5 0.8 1 0.5 0.8 1 version: 1.1 41

11.. FQuunacdtiroantsicaEnqduLaimtioitnss eLearn.Punjab eLearn.Punjab Plot the points (x, y). Connecting these points we get two straight lines, which is thegraph of a discontinuous function.Example 2: Graph the function deined by y= x2 - 9, x≠3 x - 3Solution: The domain of the function consists of all real numbers except 3.When x = 3, both the numerator and denominator are zero, and 0 is undeined. 0Simplifying we get y = x2 -9 = (x - 3)(x + 3) = x + 3 provided x ≠ 3 . x -3 x -3We prepare a table for diferent values of x and y satisfy the equation y = x + 3 and x ≠ 3. X -3 -2 -1 0 1 2 2.9 3 3.1 4 Y 0 1 2 3 4 5 5.9 6 6.1 7 Plot the points (x, y) and joining these points we getthe graph of the function which is a straight line except thepoint (3, 6). The graph is shown in the igure. This is a brokenstraight line with a break at the point (3, 6).1.7.7 Graphical Solution of the Equations (i) cos x = x (ii) sin x = x (iii) tan x = x We solve the equation cos x = x and leave the other two equations as an exercise forthe students.Solution: To ind the solution of the equation cos x = x, we draw the graphs of the two functions y = x and y = cos x : -p 7 x 7 p version: 1.1 42

11.. FQuunacdtiroantsicaEndquLaimtioitnss eLearn.Punjab eLearn.PunjabScale for graphsAlong x-axis, length of side o f small square = p radian 6Along y-axis, length of side of small square = 0.1 unitTwo points (0, 0) and ( (p/3,1) lie on the line y = x We prepare a table for some values of x and y in the interval -p 7 x 7 p it satisfying theequation y = cos x. x -p -5p/6 -2p/3 -p/2 -p/3 -p/6 0 p/6 p/3 p/2 2p/3 5p/6 py = cos x -1 -.87 -.5 0 -.5 .87 1 .87 .5 0 -.5 -.87 -1 The graph shows that the equations y = x and y = cos x intersect at only wherex = 43 p radian = 0.73 180Check: cos  43 p  = cos 43o = 0.73 180 version: 1.1 43

11.. FQuunacdtiroantsicaEnqduLaimtioitnss eLearn.Punjab eLearn.PunjabNote: Since the scales along the two axes are diferent so the line y = x is not equally inclined to both the axes. EXERCISE 1.51. Draw the graphs of the following equations(i) x2 + y2 = 9 (ii) x2 + y2 = 1 16 4(iii) y = e2x (iv) y = 3x2. Graph the curves that has the parametric equations given below(i) x = t , y = t2 , -3 7 t 7 3 where “t” is a parameter(ii) x = t -1 , y = 2t -1, -1 < t < 5(iii) x = sec q , y = tan q where “t” is a parameter where “q” is a parameter3. Draw the graphs of the functions deined below and ind whether they are continuous.(i) y = 2xx - 1 if x < 3 + 1 if x ≥ 3(ii) y= x2 - 4 x≠2 x-2(iii) y =  x+ 3 if x ≠ 3 2 if x = 3(iv) y= x2 - 16 x≠4 x-44. Find the graphical solution of the following equations:(i) x = sin 2x(ii) x = cos x 2(iii) 2x =tan x version: 1.1 44

CHAPTER version: 1.12 DIFFERENTIATION Animation 2.1: Increasing and Decreasing Functions Source and credit: eLearn.Punjab

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab eLearn.Punjab2.1 INTRODUCTION The ancient Greeks knew the concepts of area, volume and centroids etc. which arerelated to integral calculus. Later on, in the seventeenth century, Sir Isaac Newton, an Englishmathematician (1642-1727) and Gottfried Whilhelm Leibniz, a German mathematician,(1646-1716) considered the problem of instantaneous rates of change. They reachedindependently to the invention of diferential calculus. After the development of calculus,mathematics became a powerful tool for dealing with rates of change and describing thephysical universe.Dependent and Independent Variables In diferential calculus, we mainly deal with the rate of change of a dependent variablewith respect to one or more independent variables. Now, we irst explain the terms dependentand independent variables. We usually write y ∈f ( x) where f ( x) is the value of f at x Df (the domain of the functionf ). Let us consider the functional relation =v f ( x=) x2 +1....... (A) For diferent values of x ∈ Df , f ( x) or the expression x2 +1 assumes diferent values.For example; if x = 1, 1.5, 2 etc., then f (1=) (1)2 +1= 2, f (1.5=) (1.5)2 +1= 2.25 +1= 3.25 f (2) = (2)2 +1= 4 +1 = 5 We see that for the change 1.5 - 1 = 0.5 in the value of x , the corresponding change inthe value of y or f ( x) is given byf (1.5) - f (1=) 3.25 - 2= 1.25 It is obvious that the change in the value of the expression x2 +1 (or f ( x) ) dependsupon the change in the value of the variable x . As x behaves independently, so we call it theindependent variable. But the behaviour of y or f ( x) depends on the variable x , so we call itthe dependent variable. The change in the value of x (positive or negative) is called the increment of x and isdenoted by the symbol d x (read as delta x ). The corresponding change in the dependentvariable y or f ( x) for the change d x in the value of x is denoted by d y or d f = f ( x + d x) - f ( x) . version: 1.1 2

21.. DQifuaerdernattiiactEioqnuations eLearn.Punjab eLearn.PunjabUsually the small changes in the values of the variables are taken as increments of variables.Note: In this Chapter we shall discuss funcions of the form y = f(x) where xdDf and is called an independent variable while y is called the dependent variable.2.1.1 AVERAGE RATE OF CHANGE Suppose a particle (or an object) is moving in a straight line and its positions (fromsome ixed point) after times t and t1 are given by s(t ) and s (t1 ), then the distance traveled inthe time interval t1 - t where t1 > t is s(t1 ) - s(t )and the diference quotient s(t1) - s(t ) (i) t1 - trepresents the average rate of change of distance over the time interval t1 - t .If t1 - t is not small, then the average rate of change does not represent an accurate rateof change near t. We can elaborate this idea by a moving particle in a straight line whoseposition in metres after t seconds is given by s(t=) t2 + tWe construct a table for diferent values of t as under:Interval Average rate of change (i.e. average speed)t = 3 secs to t = 5 secs s(5) - s(3) = (25 + 5) - (9 + 3) = 30 -12 =9 2 5-3 2t = 3 secs to t = 4 secs s(4) - s(3) = (16 + 4) -12 = 20 -12 = 8t = 3 secs to t = 3.5 secs 1 4-3 1 s(3.5) - s(3)  49 + 7  - 12 15 4 2 3.5 - 3 = = 4 =7.5 0.5 0.5 We see that none of average rates of change approximates to the actual speed of theparticle after 3 seconds. version: 1.1 3

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab eLearn.PunjabNow we construct a table by taking small intervals. Interval Average rate of changet = 3 secs to t = 3.1 secs ( )(3.1)2 + 3.1 - 12 = 12.71 -12 = 0.71 =7.1 3.1 - 3 0.1 0.1t = 3 secs to t = 3.01 secs ( )(3.01)2 + 3.01 - 12 = 12.0701 -12 = 0.0701 =7.01 3.01 - 3 0.01 0.01t = 3 secs to t = 3.001 secs ( )(3.001)2 + 3.001 - 12 = 12.007001 -12 = 0.007001 =7.001 3.001 - 3 0.001 0.001 The above table shows that the average rate of change after 3 seconds approximatesto 7 metre/sec. as the length of the interval becomes very very small. In other words, we cansay that the speed of the particle is 7 metre/sec. after 3 seconds. If t1 = t + d tthen the diference quoteint (i) becomes s(t +dt) - s(t) dtwhich represents the average rate of change of distance over the interval d t andlim s (t + dt) - s(t )d t→0 dt , provided this limit exists, is called the instantaneous rate of changeof distance ‘s’ at time t .2.1.2 Derivative of a Function Let f be a real valued function continuous in the interval ( x,x1 ) ⊆ Df (the domain off ), thendiference quotient f ( x1) - f ( x) (i) x1 - xrepresents the average rate of change in the value of f with respect to the change x1 - x inthe value of independent variable x .If x1, approaches to x , then version: 1.1 4

21.. DQifuaerdernattiiactEioqnuations eLearn.Punjab eLearn.Punjab lim f ( x1) - f ( x) x1 → x x1 - xprovided this limit exists, is called the instantaneous rate of change of f with respect to xat x and is written as f ' ( x) .If x1 = x + d x i.e., x1 - x = d x ,then the expression (i) can be expressed as f (x +d x)- f (x) (ii) dxand lim f (x +d x)- f (x) (iii) d x→0 dxprovided the limit exists, is deined to be the derivative of f (or diferential coeicientof f ) with respect to x at x and is denoted by f '( x) (read as “f-prime of x ”). The domain off ‘consists of all x for which the limit exists. If x ∈ Df and f '( x) exists, then f is said to bediferentiable at x . The process of inding f ‘ is called diferentiation.Notation for Derivative Several notations are used for derivatives. We have used the functional symbol f ' ( x) ,for the derivative of f at x . For the function y = f ( x). y +d y = f (x +d x)- f ( ) where d y is the increment of y (change in the value of y ) corresponding to d x ,thechange in the value of x , then d y = f (x +d x)- f (x) (iv) Dividing both the sides of (iv) by d x , we get dy = f (x +d x)- f (x) (v) dx dx Taking limit of both the sides of (v) as d x → 0, we have version: 1.1 5

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab eLearn.Punjablim d y = lim f (x +d x)- f (x) (vi) d xd x→0 d x→0 dxlim d y is denoted by dy , so (vi) is written as dy = f '(x) d x dx dxd x→0Note: The symbol dy is used for the derivative of y with respect to x and here it is not a dxquotient of dy and dx. dy is also denoted by y ’. dxNow we write, in a table the notations for the derivative of y = f ( x)used by diferentmathematicians:Name of Leibniz Newton Lagrange CauchyMathematicianNotation used for derivative dy df f (x) f ' ( x) Df ( x) dx or dxIf we replace x + d x by x and x by a, then the expressionf ( x + d x) - f ( x) becomes f ( x) - f (a) . and the change d x in the independent variable, in thiscase, is x - a .So the expression f (x +d x)- f (x) is written as f (x)- f (a) (vii) dx x-aTaking the limit of the expressiom(vii) when x → a , giveslim f (x)- f (a) = f ' (a). Here f ' (a)x→a x-ais called the derivative of f at x = a . version: 1.1 6